The Cartesian equation of line $6 x - 2 = 3 y + 1 = 2 z - 2$ is given by

\frac{3 x - 1}{3} = \frac{3 y + 1}{6} = \frac{z - 1}{3}

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\frac{3 x + 1}{3} = \frac{3 y - 1}{6} = \frac{z - 1}{3}

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\frac{3 x - 1}{3} = \frac{3 y - 1}{6} = \frac{z - 1}{3}

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\frac{3 x - 1}{6} = \frac{3 y - 1}{3} = \frac{z - 1}{3}

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Solution

The correct option is A $\frac{3 x - 1}{3} = \frac{3 y + 1}{6} = \frac{z - 1}{3}$
Solution -

$6 x - 2 = 3 y + 1 = 2 z - 2$

$\frac{6 x - 2}{6} = \frac{3 y + 1}{6} = \frac{2 z - 2}{6}$

$\frac{x - \frac{1}{3}}{1} = \frac{y + \frac{1}{3}}{2} = \frac{z - 1}{3}$
DR $(1, 2, 3)$

Point passing through line $(1 / 3, - 1 / 3, 1)$

$\frac{3 x - 1}{3} = \frac{3 y + 1}{6} = \frac{z - 1}{3}$

A is correct.

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Similar questions

x + y + z = x y z

\frac{3 x - x^{3}}{1 - 3 x^{2}} + \frac{3 y - y^{3}}{1 - 3 y^{2}} + \frac{3 z - z^{3}}{1 - 3 z^{2}} = \frac{3 x - x^{3}}{1 - 3 x^{2}} \cdot \frac{3 y - y^{3}}{1 - 3 y^{2}} \cdot \frac{3 z - z^{3}}{1 - 3 z^{2}}

A = ⎡ ⎢ ⎣ \begin{matrix} 3 & 1 & 2 3 & 2 & - 3 2 & 0 & - 1 \end{matrix} ⎤ ⎥ ⎦

A^{- 1}

3 x + 3 y + 2 z = 1

x + 2 y = 4

2 x - 3 y - z = 5

3 y - 2 z - 1 = 0 = 3 x - z + 4

(2, - 1, 6)

\frac{x}{2} = \frac{y}{- 3} = \frac{z}{1}

\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{5}

A = ⎡ ⎢ ⎣ \begin{matrix} 3 & 1 & 2 3 & 2 & - 3 2 & 0 & - 1 \end{matrix} ⎤ ⎥ ⎦

A^{- 1} .

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